Dtyjlui

How does the sum
$$f(x)=sum_{n=1}^infty frac{x^n}{n^n}$$
behave asymptotically as $xtoinfty$? It appears that $f(x)$ asymptotically dominates any polynomial and is dominated by any exponential, so we might consider $log f(x)$ rather than $f(x)$.

I apologize for having no work to show on this problem; I have no idea how to begin tackling a problem regarding the asymptotics of a function given its power series (of which there is no hope of evaluating in closed form). Hopefully an answer will provide me with some tools for doing so.

Written from an airport. This is somewhat sketchy when comparing solutions to differential equations, but hopefully not too much for you to fill the gaps.

The main idea: bounding $f$ via differential partial equation.

We have
$$
f'(x) = sum_{n=1}^infty frac{x^{n-1}}{n^{n-1}}
= sum_{n=0}^infty frac{x^{n}}{(n+1)^{n}}
= 1+sum_{n=1}^infty frac{frac{x^{n}}{n^n} }{left(1+frac{1}{n}right)^{n}}
> 1+frac{1}{e}sum_{n=1}^infty frac{x^{n}}{n^n} = 1+frac{1}{e}f(x) tag{1}
$$
so in particular
$$
f' > 1+frac{1}{e}f tag{2}
$$
Since $f(0) = 0$, and the solution to $g' = 1+e^{-1}g$ with $g(0)=0$ is given by $g(x) = e^{x/e+1}-e$, we have
$$
f(x) geq e^{x/e+1}-e > 2e^{x/e} , qquad x>4tag{3}
$$
($x>4$ for the second inequality to kick in).
Now, from $(1)$ we also have
$$
f' < 1+f tag{4}
$$
(we can even improve this to $f' < 1+frac{2}{3}f$), which this time gives
$$
f(x) leq e^x - 1tag{5}
$$

Overall, for $x>4$,
$$
2e^{x/e} leq f(x) leq e^x - 1 tag{6}
$$
which provides a loose estimate of the asymptotic growth of $f$: namely, $boxed{f(x) = e^{Theta(x)}}$.

Further: Improving (slightly) on the lower bound on $log f$ by the low-order terms, and improving on the constant in the main asymptotics of the upper bound of $log f$.

I will show
$$
h(x) leq f(x) leq g(x) tag{7}
$$
where
$$
begin{align}
log h(x) &= frac{1}{e}x + 4 - logfrac{32}{3} + o(1) tag{8}\
log g(x) &= frac{256}{625}x + O(1) tag{9}
end{align}
$$
(note that $frac{256}{625} approx frac{1}{e}+0.04$). Moreover, this can be improved by the same method, pushing to more accurate estimates, but this will get uglier. (One can also push the Taylor expansion above further, based on (12) and (13). I stopped at $o(1)$).

The observation is that for the upper and lower bound, we bounded uniformly the coefficients by
$$
forall n geq 1, qquad frac{1}{n^n} leq frac{1}{left(1+frac{1}{n}right)^{n}} cdot frac{1}{n^n} leq frac{1}{e}cdot frac{1}{n^n}
$$
to obtain the two corresponding differential equations. We can do better by handling the first few terms more tightly. Namely, we have
$$
left(1+frac{1}{n}right)^n = begin{cases}
frac{1}{2} & n=1\
frac{4}{9} & n=2\
frac{27}{64} & n=3\
frac{256}{625} & n=4
end{cases}
$$
(and, of course, $left(1+frac{1}{n}right)^n$ is decreasing to $1/e$). Thus, we can leverage this and solve instead the following two differential equations for $h$ and $g$:
begin{align}
h'(x) &= 1 + left(frac{1}{2} - frac{1}{e}right) x + left(frac{4}{9} - frac{1}{e}right) frac{x^2}{4} + left( frac{27}{64} - frac{1}{e}right) frac{x^3}{27} + frac{1}{e}h(x)tag{10}\
g'(x) &= 1 + left(frac{1}{2} - frac{256}{625}right) x + left(frac{4}{9} - frac{256}{625}right) frac{x^2}{4} + left( frac{27}{64} - frac{256}{625}right) frac{x^3}{27} + frac{256}{625}g(x)tag{11}
end{align}
subject to $h(0)=g(0)=0$. Solving those gives a nasty expression,
begin{align}
h(x) &= frac{3}{32} e^{4 + frac{1}{e}x}
+ left(frac{1}{27} - frac{e}{64}right) x^3
+ left(frac{1}{4} - frac{3 e^2}{64}right) x^2
+ left(1 - frac{3e^3}{32}right) x
-frac{3e^4}{32} tag{12} \
g(x) &= frac{457763671875}{137438953472}e^{frac{256}{625}x}
- frac{491}{442368}x^3
- frac{123299}{4194304}x^2
- frac{195550963}{536870912}x
-frac{457763671875}{137438953472} tag{13} \
end{align}
leading to the claimed (8) and (9).

Below, a plot illustrating those approximations:

There is an analogue of Laplace's method which works for sums. $n ln(x/n)$ attains the maximum at $n = x/e$. Writing the exponent as $n ln(x/n) = x/e - x xi^2$, computing the expansion of $n'(xi)$ at $xi = 0$ and extending the integration range to $(-infty, infty)$, we obtain
$$frac {n'(xi)} x =
sqrt{frac 2 e} + c_1 xi -
frac 1 6 sqrt{frac e 2} ,xi^2 + c_3 xi^3 + O(xi^4),
quad xi to 0,\
sum_{n geq 1} frac {x^n} {n^n} =
int_{-infty}^infty
x left( sqrt{frac 2 e} - frac 1 6 sqrt{frac e 2} ,xi^2 +
O(xi^4) right)
e^{x/e - x xi^2} dxi = \
sqrt{frac pi 2} ,e^{x/e} left(
2sqrt{frac x e} - frac 1 {12} sqrt{frac e x} + O(x^{-3/2}) right),
quad x to infty,$$
which gives $ln f(x)$ with an error of order $O(x^{-2})$.